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Calculate the number of milliliters of 0.472 M Ba(OH)2 required to precipitate all of the Ca2+ ions in 197 mL of 0.671 M CaBr2 solution as Ca(OH)2. The equation for the reaction is: CaBr2(aq) + Ba(OH)2(aq) Ca(OH)2(s) + BaBr2(aq)

Respuesta :

Explanation:

[tex] \sqrt[ log_{ \tan( \geqslant e | \binom{?}{?} | \times \frac{0.671 log( \gamma ) }{38019} ) }( {2.8m}^{e} ) ]{197 {6}^{2} } = > 6.819t[/tex]

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